如何对第一个值（字符串）进行分组并添加相同日期的值？

Question

JAVASCRIPT-JQUERY求和如何对第一个值（日期）进行分组并添加相同日期的值？

ARRAY：

0: (5) ["11-2019", 0, 20, 0, 0]
1: (5) ["11-2019", 41, 0, 0, 0]
2: (5) ["11-2019", 0, 0, 29, 0]
3: (5) ["11-2019", 0, 0, 0, 60]
4: (5) ["09-2019", 0, 1, 0, 0]
5: (5) ["09-2019", 0, 0, 1, 0]
6: (5) ["09-2019", 0, 0, 0, 1]
7: (5) ["05-2019", 2, 0, 0, 0]

OUT：

0: (5) ["11-2019", 41, 20, 29, 60]
1: (5) ["09-2019", 0, 1, 1, 1]
2: (5) ["05-2019", 2, 0, 0, 0]

---

result = DataAll.reduce(function(r, a) {
     a.forEach(function(b, i) {

         r[i] = (r[i] || 0) + b;
         console.log(r[i]);
     });
     return r;
 }, []);

Answer 1

我将在结果集中找到该数组并更新所有值。

var data = [["11-2019", 0, 20, 0, 0], ["11-2019", 41, 0, 0, 0], ["11-2019", 0, 0, 29, 0], ["11-2019", 0, 0, 0, 60], ["09-2019", 0, 1, 0, 0], ["09-2019", 0, 0, 1, 0], ["09-2019", 0, 0, 0, 1], ["05-2019", 2, 0, 0, 0]],
    result = data.reduce((r, a) => {
        var temp = r.find(([date]) => date === a[0])
        if (temp) {
            for (var i = 1; i < a.length; i++) temp[i] += a[i];
        } else {
            r.push([...a]);
        }
        return r;
    }, []);

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

我在脚本中添加了filter，以从结果中删除0值。如果您确实要使用0值，请使用

acc[curr[0]]=(acc[curr[0]]||[]).concat(curr.slice(1));

代替。

var inp=[["11-2019", 0, 20, 0, 0],
 ["11-2019", 41, 0, 0, 0],
 ["11-2019", 0, 0, 29, 0],
 ["11-2019", 0, 0, 0, 60],
 ["09-2019", 0, 1, 0, 0],
 ["09-2019", 0, 0, 1, 0],
 ["09-2019", 0, 0, 0, 1],
 ["05-2019", 2, 0, 0, 0]];
 var out=inp.reduce((acc,curr)=>{
   acc[curr[0]]=(acc[curr[0]]||[]).concat(curr.slice(1).filter(v=>v>0));
   return acc
 }, {});
  console.log(out);
  
  // and to get it into your format:
  var outarr=Object.keys(out).map(k=>[k].concat(out[k]))
  console.log(outarr)

对，如果您想要sum，那么我的版本如下。感谢Nina首先提供正确答案。 ;-）

var inp=[["11-2019", 0, 20, 0, 0],
 ["11-2019", 41, 0, 0, 0],
 ["11-2019", 0, 0, 29, 0],
 ["11-2019", 0, 0, 0, 60],
 ["09-2019", 0, 1, 0, 0],
 ["09-2019", 0, 0, 1, 0],
 ["09-2019", 0, 0, 0, 1],
 ["05-2019", 2, 0, 0, 0]];
 
 let out=inp.reduce((acc,cur)=>{
   if(acc[cur[0]]) acc[cur[0]].forEach((v,i,a)=>a[i]+=cur[i+1]);
   else acc[cur[0]]=cur.slice(1)
   return acc
 }, {} );
 outarr=Object.keys(out).map(k=>[k].concat(out[k]))
 
 console.log(outarr)